Question

设数列｛an｝的前n项和Sn＝n²．．｛bn｝是各项均为正数的等比数列且a1＝b1 a5×b3＝1

球1数列｛an｝和｛bn｝通项公式
2．设cn=an×bn求数列｛cn｝的前n项和Tn

Answer 1

n=1,时a1=S1=1
n>=2时,an=Sn-S(n-1)=n^2-(n-1)^2=2n-1
故an=2n-1
a5=9
b1=a1=1
a5*b3=1,b3=1/9
q^2=b3/b1=1/9
q=1/3
故bn=b1q^(n-1)=(1/3)^(n-1)
(2)cn=(2n-1)*(1/3)^(n-1)=3(2n-1)*(1/3)^n
Tn=3[1*1/3+3*(1/3)^2+...+(2n-1)*(1/3)^n]
1/3Tn=3[1*(1/3)^2+3*(1/3)^3+...+(2n-1)*(1/3)^(n+1)]
Tn-1/3Tn=3[1/3+2(1/3)^2+2(1/3)^3+...+2(1/3)^n-(2n-1)*(1/3)^(n+1)]
2Tn/3=3[1/3+2*(1/3)^2*(1-(1/3)^(n-1))/(1-1/3)-(2n-1)*(1/3)^(n+1)]
Tn=9/2[1/3+2/9(1-(1/3)^n*3)*3/2-(2n-1)*(1/3)^(n+1)]
=3/2+3/2-9/2*(1/3)^n-(2n-1)*(1/3)^(n)*1/3*9/2
=3-9/2*(1/3)^n-3/2*(2n-1)*(1/3)^n